JEE Main 2024 — 4 April, Shift 2 — Question 17

$\cos ^{-1} \mathrm{x}-\left(\frac{\pi}{2}-\cos ^{-1} \mathrm{y}\right)=\alpha$

$\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha$

$\alpha \in\left[-\frac{\pi}{2}, \pi\right], \frac{\pi}{2}+\alpha \in\left[0, \frac{3 \pi}{2}\right]$

$\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)=\frac{\pi}{2}+\alpha$

$x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}=-\sin \alpha$

$(x y+\sin \alpha)=\left(1-x^{2}\right)\left(1-y^{2}\right)$

$x^{2} y^{2}+2 x y \sin a+\sin ^{2} a=1-x^{2}-y^{2}+x^{2} y^{2}$

$x^{2}+y^{2}+2 x y \sin \alpha=1-\sin ^{2} \alpha$

$x^{2}+y^{2}+2 x y \sin \alpha=\cos ^{2} \alpha$

Min. value of $\cos ^{2} \alpha=0$

At $\alpha=\frac{\pi}{2}$